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Thread: Initial Value Problems

  1. #1
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    Initial Value Problems

    It's been awhile since I have had calculus and dealing with differential equations and initial value problems. Looking for some help with some easy problems.

    Find the solution of the given initial value problem.

    (1) $\displaystyle y' + 2y = te^{-2t}$, $\displaystyle y(1)=0$

    (2) $\displaystyle t^3y' + 4t^2y = e^{-t}$, $\displaystyle y(-1)=0, t < 0$

    A walkthrough of these would be greatly appreciated. Thanks in advance
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  2. #2
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    Quote Originally Posted by Proof_of_life View Post
    (1) $\displaystyle y' + 2y = te^{-2t}$, $\displaystyle y(1)=0$
    The integrating factor is $\displaystyle f(x) = e^{\int 2 dx}= e^{2x}$.
    Therefore the solution is,
    $\displaystyle y=\frac{1}{f(x)} \int f(x) xe^{-2x} dx = e^{-2x} \int x dx = \frac{1}{2}x^2e^{-2x} + ke^{-2x}$.
    Now $\displaystyle y(1) = \frac{1}{2}e^{-2} + k e^{-2} = 0 \implies k = -\frac{1}{2}$.
    Thus, $\displaystyle y = \frac{1}{2}x^2e^{-2x} - \frac{1}{2} e^{-2x}$.

    (2) $\displaystyle t^3y' + 4t^2y = e^{-t}$, $\displaystyle y(-1)=0, t < 0$

    A walkthrough of these would be greatly appreciated. Thanks in advance
    This one is similar. Just divide out the leading coefficient first.
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  3. #3
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    I was about to reply that I figured out both of these. Thanks for replying though!
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